Calculations:Transformer Efficiency and Regulation

Transformer Efficiency and Regulation

Introduction

The calculation of efficiency differs from IEC 60076 and ANSI C57.12 Standards because of the difference in defining rated kVA by these standards.

IEC 60076 Standard and ANSI C57.12 Standard

The rated power is defined by IEC 60076 as “When the transformer has rated voltage applied to a primary winding, and rated current flows through the terminals of a secondary winding the trans- former receives the relevant rated power for that pair of windings”.

This implies that it is a value of apparent power input to the transformer, including its own absorption of active and reactive power. The apparent power that the transformer delivers to the circuit connected to the terminals of the secondary winding under rated loading differs from the rated power. The voltage across the secondary terminals differs from the rated voltage by the voltage drop (or rise) in the transformer.

This is different from the definition as per ANSI C57.12.00 standard, where the rated kVA is defined as

The rated kVA of a transformer shall be the output that can be delivered for the time specified at rated secondary voltage and rated frequency without exceeding the specified temperature rise limitations under prescribed conditions of test, and within the limitations of established standards.

Calculation as per ANSI Standard

The efficiency of a transformer is the ratio of its useful power output to its total power input.

${\eta }\%={\frac {100P_{o}}{P_{i}}}={\frac {P_{i}-P_{L}}{P_{i}}}=\left(1-{\frac {P_{L}}{P_{o}+P_{L}}}\right)\times 100\,$

${\eta }=100\times \left(1-{\frac {P_{L}}{P_{o}+P_{L}}}\right)\,$

where

${\eta }\%=\,$ efficiency (%)

${P_{o}}=\,$ output

${P_{i}}=\,$ input

${P_{L}L}=\,$ losses

For power factor of "S" and load of "x" per unit, the formula for efficiency is

${\eta }\%=100\times \left(1-{\frac {W_{i}+x^{2}W_{c}}{xSP_{o}+W_{1}+x^{2}W_{c}}}\right)\,$

where

${\eta }\%=\,$ efficiency (%)

${W_{i}}=\,$ no-load loss (W)

${W_{c}}=\,$ load loss at full load at reference temperature (W)

${x}=\,$ per unit load

${S}=\,$ power factor per unit

The calculation of efficiency differs from IEC 60076 and ANSI C57.12 Standards because of the difference in defining rated kVA by these standards.

Calculation as per IEC Standard

The voltage drop (or rise) of the transformer secondary will have to be considered, while the efficiency is calculated, as per IEC standard.

${\eta }\%=100-\left({\frac {P_{o}+n^{2}P_{k}}{nS_{n}cos\phi \left(1-{\frac {U_{\phi n}}{100}}\right)+P_{o}+n^{2}P_{k}}}\right)\times 100\,$

where

${\eta }\%=\,$ efficiency (%)

${P_{o}}=\,$ no-load loss (kW)

${n}={\frac {Load}{RatedPower}}\,$ (kVA)

${P_{k}}=\,$ load loss (kW)

${cos\phi }=\,$ power factor

${U_{\phi n}}=\,$ voltage drop (or rise) %

${U_{\phi n}}=n\times \left(rcos\phi +xsin\phi \right)+{\frac {\left[n\times \left(rsin\phi -xcos\phi \right)\right]^{2}}{200}}\,$

Example Calculations

An example for a typical power transformer.

Transformer parameters

kVA – 1500

Voltage ratio – 11/0.433

% Impedance – 6.0

% Resistance – 0.6

% Reactance – 5.97

No-load loss – 2.3 kW

Load loss – 9.0 kW

Transformer Impedance

$Z=0.6+j5.97=6.0\Omega \,$

Efficiency at 100% load unity power factor

At unity power factor (full load)

$rcos\phi +xsin\phi =0.6\times 1+5.97\times 0=0.6\,$

$rsin\phi -xsin\phi =0.6\times 0-5.97\times 1=-5.97\,$

Voltage Drop$=0.6+{\frac {-5.97^{2}}{200}}=0.778\%\,$

As per IEC Standard

${\eta }\%=100-\left({\frac {P_{o}+n^{2}P_{k}}{nS_{n}cos\phi \left(1-{\frac {U_{\phi n}}{100}}\right)+P_{o}+n^{2}P_{k}}}\right)\times 100\,$

${\eta }\%=100-\left({\frac {2.3+9.0}{1500\times 1\left(1-{\frac {0.778}{100}}\right)+2.3+9.0}}\right)\times 100\,$

${\eta }\%=100-\left({\frac {11.3}{1499.63}}\right)\times 100\,$

${\eta }\%=99.246\%\,$

As per ANSI Standard

${\eta }\%=100\times \left(1-{\frac {W_{i}+x^{2}W_{c}}{xSP_{o}+W_{1}+x^{2}W_{c}}}\right)\,$

${\eta }\%=100\times \left(1-{\frac {2.3+9.0}{1500+2.3+9.0}}\right)\,$

${\eta }\%=99.252\%\,$

Efficiency at 100% load 0.8 power factor

At 0.8 power factor (full load)

$rcos\phi +xsin\phi =0.6\times 0.8+5.97\times 0.6=4.062\,$

$rsin\phi -xsin\phi =0.6\times 0.6-5.97\times 0.8=-4.416\,$

Voltage Drop$=4.062+{\frac {-4.416^{2}}{200}}=4.159\%\,$

As per IEC Standard

${\eta }\%=100-\left({\frac {P_{o}+n^{2}P_{k}}{nS_{n}cos\phi \left(1-{\frac {U_{\phi n}}{100}}\right)+P_{o}+n^{2}P_{k}}}\right)\times 100\,$

${\eta }\%=100-\left({\frac {2.3+9.0}{1500\times 0.8\left(1-{\frac {4.159}{100}}\right)+2.3+9.0}}\right)\times 100\,$

${\eta }\%=100-\left({\frac {11.3}{1161.39}}\right)\times 100\,$

${\eta }\%=99.027\%\,$

As per ANSI Standard

${\eta }\%=100\times \left(1-{\frac {W_{i}+x^{2}W_{c}}{xSP_{o}+W_{1}+x^{2}W_{c}}}\right)\,$

${\eta }\%=100\times \left(1-{\frac {2.3+9.0}{1500\times 0.8+2.3+9.0}}\right)\,$

${\eta }\%=99.067\%\,$

Efficiency at 50% load unity power factor

At unity power factor (50% load)

Voltage Drop$=0.5\times 0.6+{\frac {(-5.97\times 0.5)^{2}}{200}}=0.3445\%\,$

At 0.8 power factor (50% load)

Voltage Drop$=0.5\times 4.062+{\frac {(-4.416\times 0.5)^{2}}{200}}=2.055\%\,$